HDU 1081 To The Max DP

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题意：给你一个矩阵，求最大子矩阵的和。

思路：首先a[i][j]表示第i行的前缀和。枚举i,j列,dp[i][j][k]代表从i列到j列，前k行所能形成的最大子矩阵。

DP方程：dp[i][j][k]=max(dp[i][j][k]+a[k][j]−a[k][i−1],a[k][j]−a[k][i−1])

http://acm.hdu.edu.cn/showproblem.php?pid=1081

/*********************************************    Problem : HDU 1081    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = a ; i <= b ; i ++)#define rrep(i,a,b) for(int i = b ; i >= a ; i --)#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;const int MAXN = 1e5;const int MAXE = 2e5;typedef long long LL;typedef unsigned long long ULL;int T,n,m,k;int a[105][105];int dp[105][105][105];void init() {}void input() {    rep(i,1,n) rep(j,1,n) scanf("%d",&a[i][j]) , a[i][j] += a[i][j-1];//第i行,j的前缀和。}void solve() {    int ans = -INF;    //rep(i,0,n) rep(j,0,n) rep(k,0,n) dp[i][j][k] = -INF;    rep(i,1,n) rep(j,i,n) rep(k,1,n){        //代表i列到第j列。以k行结尾的最大子矩阵值。        dp[i][j][k] = max(dp[i][j][k-1]+a[k][j]-a[k][i-1],a[k][j]-a[k][i-1]);        ans = max(ans,dp[i][j][k]);    }    printf("%d\n",ans);}int main(void) {    //freopen("a.in","r",stdin);    //scanf("%d",&T);    //while(T--) {    init();    while(~scanf("%d",&n)) {        input();        solve();    }    return 0;}

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